# What is the antiderivative of xe^x?

Jul 8, 2016

Let $u = x$ and $\mathrm{dv} = {e}^{x} \mathrm{dx}$, so $\mathrm{du} = \mathrm{dx}$ and $v = {e}^{x}$

Knowing that

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

You can see that

$\int x {e}^{x} \mathrm{dx}$
$= x {e}^{x} - \int {e}^{x} \mathrm{dx}$
$= x {e}^{x} - {e}^{x} + C$

$= {e}^{x} \left(x - 1\right) + C$

#### Explanation:

In this case, you'd have to perform integration by parts, which is given by the following formula:

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

Note that if you would have chosen $u = {e}^{x}$ and $\mathrm{dv} = x \mathrm{dx}$, your integral would be even more complicated.